College Trigonometry, 2011a

10.7 Trigonometric Equations and Inequalities 873
−1
(−) 0
(+)
0 1
y = π 2 − 4 arccos 2 (x)
6. To begin, we rewrite 4 arccot(3x) >πas 4 arccot(3x)−π >0. We let f(x) = 4 arccot(3x)−π,
and note the domain of f isallrealnumbers,(−∞, ∞). To find the zeros of f, we set
f(x) = 4 arccot(3x) − π = 0 and solve. We get arccot(3x) = π 4 , and since π 4
is in the range of
arccotangent, we may apply Theorem 10.27 and solve
arccot(3x) = π 4
cot(arccot(3x)) = cot ( )
π
4
3x = 1 Since cot(arccot(u)) = u.
x = 1 3
Next, we make a sign diagram for f. Since the domain of f is all real numbers, and there is
only one zero of f, x = 1 3 , we have two test intervals, ( −∞, 1 (
3)
and
1
3 , ∞) . Ideally, we wish
to find test values x in these intervals so that arccot(4x) corresponds to one of our oft-used
‘common’ angles. After a bit of computation, 18 we choose x =0forx< 1 3 and for x> 1 3 ,we
√ ( √3 )
choose x = 3
3 . We find f(0) = π>0andf 3
= − π 3
< 0. Since we are looking for where
f(x) = 4 arccot(3x) − π>0, we get our answer ( −∞, 1 3)
. To check graphically, we use the
technique in number 2c of Example 10.6.5 in Section 10.6 to graph y = 4 arccot(3x) andwe
see it is above the horizontal line y = π on ( −∞, 1 ( )
3)
= −∞, 0.3 .
(+) 0
1
3
(−)
y = 4 arccot(3x) andy = π
18 Set 3x equal to the cotangents of the ‘common angles’ and choose accordingly.